Two cars, a porsche boxster and a toyota scion xb, are traveling in the same direction, although the boxster is 115.0 m behind the scion. the speed of the boxster is 26.0 m/s and the speed of the scion is 18.9 m/s. how much time does it take for the boxster to catch the scion? [hint: what must be true about the displacement of the two cars when they meet? ]

16.197 seconds

Step-by-step explanation:

the time and the final place fot both cars is the same.

for the Boxster we have

X = Vboxster . T

being

X the distancance from the initial place of the Boxter to the meeting point

Vboxster the speed of the Boxster (26m/s)

T the time

and for the Scion

X = Xo + Vscion . T

Being

Xo the initial point of the Scion (115m)

X the distancance from the initial place of the Boxster to the meeting point

Vscion the speed of the Scion (18.9 m/s)

T the time

                 Xo + Vscion . T  = Vboxster . T

                                    Xo   = Vboxster . T - Vscion . T

                                   Xo   = (Vboxster  - Vscion) . T

Xo/(Vboxster  - Vscion)   =  T

    115m/ (26-18.9) m.s-1 = T

                          16.197 s = T

a)t = 16.2s  : time it takes for the Boxster to catch the Scion

b)The displacement of the Boxster (db) is 115 m more than the displacement of the Scion(ds)

db =ds+115

Step-by-step explanation:

Conceptual analysis

We apply the formula for constant speed movement:

v= d/t Formula (1)

v = speed in m/s

d: distance in m

t: time in s

Problem development

The time of Scion (ts) is equal time of Boxster (tb)

t (s) =tb=t

The displacement of the Boxster is 115 m more than the displacement of the Scion

db =ds+115

we apply formula (1 )car kinematics  :

Scion kinematics

18.9=ds/t

t =ds /18.9 Equation (1)

Boxster kinematics

26=db/t

26=(ds+115)/t

t=(ds+115)/26 Equation (2)

Equation (1) = Equation (2)

ds /18.9 =(ds+115)/26

18.9(ds+115)= 26 ds

18.9ds+18.9*115=26 ds

2173.5= 26 ds-18.9ds

2173.5=7.1ds

ds =2173.5÷7.1

ds=306.12m

We replace ds=306.12m in the equation (1)

t =306.12÷18.9

t = 16.2s

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

44

step-by-step explanation: